$1/a)(2x+1)^2-(2x-1)(5+2x)$
$=4x^2+4x+1-(10x+4x^2-5-2x)$
$=4x^2+4x+1-(4x^2+8x-5)$
$=4x^2-4x^2+4x-8x+1+5$
$=-4x+6$
$b)(3-x)^2+(3x-1)^2+4$
$=9-6x+x^2+9x^2-6x+1+4$
$=10x^2-12x+14$
$c)4(2x-5)(5+x)-7(4x+1)^2$
$=4(10x+2x^2-25-5x)-7(16x^2+8x+1)$
$=4(2x^2+5x-25)-112x^2-56x-7$
$=8x^2+20x-100-112x^2-56x-7$
$=-104x^2-36x-107$
$d)(x+2)^3+(7-2x)^3+5$
$=x^3+6x^2+12x+8+343−294x+84x^2−8x^3+5$
$=−7x^3+90x^2−282x+356$
$2)(x+8)(x+6)-x^2=104$
$⇔x^2+14x+48-x^2=104$
$⇔14x=56$
$⇔x=4$
Vậy $S=\{4\}$
$b)(x+1)(x+2)-(x-3)(x+4)=6$
$⇔x^2+3x+2-x^2-x+12=6$
$⇔2x=-8$
$⇔x=-4$
Vậy $S=\{-4\}$
$c)(2x+1)^2-4(x+2)^2=9$
$⇔4x^2+4x+1-4x^2-16x-16=9$
$⇔-12x=24$
$⇔x=-2$
Vậy $S=\{-2\}$
$d)(x+1)^3-x^2(x+3)=2$
$⇔x^3+3x^2+3x+1-x^3-3x^2=2$
$⇔3x=1$
$⇔x=\dfrac{1}{3}$
Vậy $S=\{\dfrac{1}{3}\}$
